Question

How do you find the linearization L(x) of f(x) = 3x^2+5x+12 at x=3?

What are the steps for this?

Answer 1

We need to find the tangent line of f at x=3

Answer 2

\[f'(x)=6x+15\]
so to find the slope use f'
and then we also know a point on the line
\[(3,3(3)^2+5(3)+12=27+15+12=54)\]

Answer 3

\[y=mx+b => 54=23(3)+b \text{ solve for b}\]

Answer 4

Wouldn't it be f'(x) = 6x + 17, when you take the derivitive?

Answer 5

oops i put an extra 1 in there lol

Answer 6

\[f'(x)=3(2x)+5=6x+5\]

Answer 7

\[(x^2)'=2x;(5x)'=5;(constant)'=0\]

Answer 8

Ok sounds good, also have another quick question, Where did you get the 23 from... on
y=mx+b=>54=23(3)+b solve for b

Answer 9

well 23 was what i found when i plug into f'
you know what i got earlier for f' lol

Answer 10

\[f'(3)=6(3)+5=18+5=23\]

Answer 11

f'(3) is the slope of the curve at x=3